Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
I hope it is no too late
Explanation:
hmmm,
In a gas, for example, the molecules are traveling in random directions at a variety of speeds - some are fast and some are slow. ... If more energy is put into the system, the average speed of the molecules will increase and more thermal energy or heat will be produced.
Answer:
Explanation:
a) Magnification = image height / object height = -9 / 18 = -0.5
b) Magnification = - image distance / object distance = -0.5
so image distance = 0.5 object distance
1/focal length = 1/image distance + 1/object distance
1/6 = 1/(0.5 object distance) + 1/object distance
object distance = 18.0 cm
c) Image appears behind the lens.
